The ages of three siblings combined is 27. The oldest is twice the age of the youngest. The middle child is 3 years older than t
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The total surface area of a coin can be solve by using this formula
Surface Area = Area Front + Area Back + Area Side
Surface Area = 2Area Front/Back + Area Side
The Area of the Front and Back of the coin can be solve by using the formula of a circle which is
And the Area of a Side of the coin can be solve by using the formula
A = πd x thickness
We will solve first the surface area of a $5 coin.
note: 2.5 cm = 25 mm
Surface Area = 2(π(d/2)^2) + πd x thickness
= 2(π(12.5)^2) + 25π x 3
= 2 (156.25π) + 75π
=312.50π + 75π
= 387.50π
Surface Area of $5 coin is 387.50π sq mm
We will next solve for surface area of a 50 cents.
note: 2 cm = 20 mm
Surface Area = 2(π(d/2)^2) + πd x thickness
= 2(π(10)^2) + 20π x 1.5
= 2 (100π) + 30π
= 200π + 30π
= 230π
The surface area of a 50 cents is 230π sq mm.
Therefore, the ratio of the total surface area of $5 coin to 50 cents is
387.50π sq mm : 230π sq mm
Surface Area = Area Front + Area Back + Area Side
Surface Area = 2Area Front/Back + Area Side
The Area of the Front and Back of the coin can be solve by using the formula of a circle which is
And the Area of a Side of the coin can be solve by using the formula
A = πd x thickness
We will solve first the surface area of a $5 coin.
note: 2.5 cm = 25 mm
Surface Area = 2(π(d/2)^2) + πd x thickness
= 2(π(12.5)^2) + 25π x 3
= 2 (156.25π) + 75π
=312.50π + 75π
= 387.50π
Surface Area of $5 coin is 387.50π sq mm
We will next solve for surface area of a 50 cents.
note: 2 cm = 20 mm
Surface Area = 2(π(d/2)^2) + πd x thickness
= 2(π(10)^2) + 20π x 1.5
= 2 (100π) + 30π
= 200π + 30π
= 230π
The surface area of a 50 cents is 230π sq mm.
Therefore, the ratio of the total surface area of $5 coin to 50 cents is
387.50π sq mm : 230π sq mm